You may use a computer algebra system to help you with the integration when finding a Fourier series, but you must show the essential steps to receive full credit.
We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_t = (2XT)_{xx}$, or $XT' = 2X''T.$ Rearranging,
$$\frac{T'}{2T} = \frac{X''}{X}.$$
For this equation to be true, both sides must equal a constant, which we denote $-\lambda.$ Thus, $\frac{T'}{2T} = -\lambda$ and $\frac{X''}{X} = -\lambda$. These two ordinary differential equations are easily solved:
while $\int_{0}^{4} x\sin\left(\frac{n\pi x}{4}\right)dx$ can be evaluated by parts by letting $u = x$, $du = dx$, $v = -\frac{4}{n\pi}\cos\left(\frac{n\pi x}{4}\right)$, and $dv = \sin\left(\frac{n\pi x}{4}\right)$:
And $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{x\pi x}{4}\right)$$ should be $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{n\pi x}{4}\right).$$
Comments
We start by separating variables, and assume $u(x,t) = X(x)T(t).$ Our partial differential equation then becomes $(XT)_t = (2XT)_{xx}$, or $XT' = 2X''T.$ Rearranging,
$$\frac{T'}{2T} = \frac{X''}{X}.$$
For this equation to be true, both sides must equal a constant, which we denote $-\lambda.$ Thus, $\frac{T'}{2T} = -\lambda$ and $\frac{X''}{X} = -\lambda$. These two ordinary differential equations are easily solved:
$$T' = -2\lambda T,$$
so
$$T = e^{-2\lambda t}.$$
Meanwhile,
$$X'' = -\lambda X,$$
so
$$X = a\cos\left(\sqrt{\lambda}x\right) + b\sin\left(\sqrt{\lambda}x\right).$$
From our initial conditions, we know
$$X(0) = a + 0 = 0,$$
so $a = 0.$ We also know
$$X(4) = b\sin\left(\sqrt{\lambda}4\right) = 0,$$
so
$$\sqrt{\lambda} = \frac{n\pi}{4}.$$
Thus, our full solution is the Fourier sine series
$$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{x\pi x}{4}\right),$$
where
$$b_n = \frac{2}{L}\int_{0}^{L} u_0(x)\sin\left(\frac{n\pi x}{L}\right)dx = \frac{1}{2}\int_{0}^{4} (4-x)\sin\left(\frac{n\pi x}{4}\right)dx = \frac{1}{2}(\int_{0}^{4} 4\sin\left(\frac{n\pi x}{4}\right)dx + \int_{0}^{4} x\sin\left(\frac{n\pi x}{4}\right)dx).$$
We tackle each integral separately. First,
$$4\int_{0}^{4} \sin\left(\frac{n\pi x}{4}\right)dx = -\frac{16}{n\pi}\cos\left(\frac{n\pi x}{4}\right) Bigr|_{0}^{4} = \frac{16}{n\pi}(1 - \cos(n\pi)) = \frac{16}{n\pi}(1-(-1)^n),$$
while $\int_{0}^{4} x\sin\left(\frac{n\pi x}{4}\right)dx$ can be evaluated by parts by letting $u = x$, $du = dx$, $v = -\frac{4}{n\pi}\cos\left(\frac{n\pi x}{4}\right)$, and $dv = \sin\left(\frac{n\pi x}{4}\right)$:
$$\int_{0}^{4} x\sin\left(\frac{n\pi x}{4}\right)dx = -\frac{4}{n\pi}x\cos\left(\frac{n\pi x}{4}\right)\Bigr|_{0}^{4} + \frac{4}{n\pi}\int_{0}^{4}\cos\left(\frac{n\pi x}{4}\right)dx = \left(-\frac{16}{n\pi}\cos(n\pi) - 0\right) + (0 - 0) = -\frac{16}{n\pi}(-1)^n. $$
Therefore,
$$b_n = \frac{1}{2}\int_{0}^{4} (4-x)\sin\left(\frac{n\pi x}{4}\right)dx = \frac{16}{n\pi}(1 - (-1)^n) - (-\frac{16}{n\pi}(-1)^n) = \frac{8}{n\pi}.$$
Thus,
$$u(x,t) = \frac{8}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}e^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{n\pi x}{4}\right).$$
Whoops, the middle term in the penultimate line should be multiplied by $\frac{1}{2}$.
And $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{x\pi x}{4}\right)$$ should be $$u(x,t) = \sum_{n=1}^{\infty}b_ne^{-\frac{n^2\pi^2}{8}t}\sin\left(\frac{n\pi x}{4}\right).$$